#2. Max./Min. on [-2,2]x[-1,1]

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Originally Posted by

**dmbocci** 2.Find the absolute extremes of $\displaystyle F(x,y)=x^2+xy, \ R={|x|\le 2, |y|\le 1}$

Someone help me set this up?

Thanks

...

So for #2, I got critical points of (0,0) (0,1) (0,-1) (0,-2) (2,1) (2,0). But do I disregard (0,-1)&(0,-2) since they are outside my boundary?

Either way, the values I got are 0,4,6. So would 0 be my minimum and if so in my answer would I say that the function has absolute min. at both (0,1) and (0,0). Or would 4 be my min at (2,0)?

And 6 would be my abs. max. at (2,1)?

Does this look right?

How did you find the critical points?

The point (0, -1) is not outside your boundary. $\displaystyle \displaystyle R$ is a rectangular region with verticies (2,1), (2,-1), -2, -1), and (-2, 1).

#2. Max./Min. on [-2,2]x[-1,1]

Quote:

Originally Posted by

**dmbocci** 2.Find the absolute extremes of $\displaystyle F(x,y)=x^2+xy,\ R=\{|x|\le2,\ |y|\le1\}$

Someone help me set this up?

Thanks

...

Quote:

Originally Posted by

**dmbocci** Not thinking on #2, whoops.

So idk if this is right, but now I have (-2,1)(-2,-1)(2,-1)(2,1)(-2,2)(2,-2)(1/2,-1)(-1/2,1) as my points to check. Does this look correct?

If so, my values output are 6,2,-1/2, and for (-2,2)&(2,-2)= 0 . But I would exclude those last two values?

Someone check my work?

How do we find the extrema for $\displaystyle \displaystyle F(x,\ y)$ over some bounded closed region, $\displaystyle \displaystyle R\ ?$ Yes, look for critical points.

Critical points occur at locations within the interior of $\displaystyle \displaystyle R,$ where $\displaystyle \displaystyle {{\partial F}\over {\partial x}}=0$ and $\displaystyle \displaystyle {{\partial F}\over {\partial y}}=0$, or where these partial derivatives do not exist. They also occur along the boundary of $\displaystyle \displaystyle R\$, where directional derivatives along that boundary are zero, or don't exist. Take all of that into account, and find the minimum & maximum values of $\displaystyle \displaystyle F$ at all of the critical points.

$\displaystyle \displaystyle {{\partial F}\over {\partial x}}=2x+y$. Setting this equal to zero gives: $\displaystyle \displaystyle y=-2x$. $\displaystyle \displaystyle {{\partial F}\over {\partial y}}=x$. Setting this equal to zero gives: $\displaystyle \displaystyle x=0$. This gives a critical point at the origin.

On the boundary $\displaystyle \displaystyle y=1$: $\displaystyle \displaystyle F(x,\ 1)=x^2+x$, so $\displaystyle \displaystyle {{dF}\over {dx}}=2x+1=0\ \ \implies\ \ x=-{{1}\over{2}}$. So, $\displaystyle \left(-{{1}\over{2}},\ 1\right)$ is a critical point. Similarly, $\displaystyle \left({{1}\over{2}},\ -1\right)$ is also a critical point.

On the boundary $\displaystyle \displaystyle x=2$: $\displaystyle \displaystyle F(2,\ y)=4+2y$, so $\displaystyle \displaystyle {{dF}\over {dy}}=2\ne0$. So there are no critical points here, except at the end points.

Critical points are thus: $\displaystyle (0,\ 0), \left(-{{1}\over{2}},\ 1\right),\ \left({{1}\over{2}},\ -1\right),\ (2,\ 1), \ (2,\ -1),\ (-2,\ -1),\ (-2,\ 1)$

Evaluate $\displaystyle \displaystyle F(x,\ y)$ at each of these points to find the extrema.